\(\dfrac{8^3.9^4-\left(-2\right)^8.81^2}{16^2.81^2+4^4.27^3}\)
giúp em với mng =(
K
Khách
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TN
1
Những câu hỏi liên quan
26 tháng 10 2021
\(A=\dfrac{\left(-3\right)^{45}\cdot5^3\cdot2^{12}}{5^4\cdot3^{44}\cdot\left(-2\right)^{11}}=\dfrac{\left(-3\right)^{45}\cdot\left(-2\right)^{12}}{5\cdot\left(-3\right)^{44}\cdot\left(-2\right)^{11}}=\dfrac{\left(-3\right)\cdot\left(-2\right)}{5}=\dfrac{6}{5}\)
AH
Akai Haruma
Giáo viên
11 tháng 8 2021
Câu 1,2 bạn đã đăng và có lời giải rồi
Câu 3:
\(=\frac{(\sqrt{3})^2+(2\sqrt{5})^2-2.\sqrt{3}.2\sqrt{5}}{\sqrt{2}(\sqrt{3}-2\sqrt{5})}=\frac{(\sqrt{3}-2\sqrt{5})^2}{\sqrt{2}(\sqrt{3}-2\sqrt{5})}=\frac{\sqrt{3}-2\sqrt{5}}{\sqrt{2}}\)
L
1
9 tháng 7 2018
\(\dfrac{4^3\cdot9^5\cdot\left(-2\right)^6}{16^4\cdot\left(-27\right)^2}\)
=\(\dfrac{\left(2^2\right)^3\cdot\left(3^2\right)^5\cdot2^6}{\left(2^4\right)^4\cdot\left(-3^3\right)^2}\)
=\(\dfrac{2^6\cdot3^{10}\cdot2^6}{2^{16}\cdot\left(-3\right)^6}\)
=\(\dfrac{2^{6+6}\cdot3^{10}}{2^{16}\cdot3^6}\)
=\(\dfrac{2^{12}\cdot3^{10}}{2^{16}\cdot3^6}\)
=\(\dfrac{3^4}{2^4}\)
=\(\dfrac{81}{16}\)
HY
1
AH
Akai Haruma
Giáo viên
5 tháng 2
Lời giải:
\(M=\frac{9^4.27^5.3^6.3^4}{3^8.81^4.234.8^2}=\frac{(3^2)^4.(3^3)^5.3^6.3^4}{3^8.(3^4)^4.2.3^2.13.(2^3)^2}\)
\(=\frac{3^8.3^{15}.3^6.3^4}{3^8.3^{16}.2.3^2.13.2^6}=\frac{3^{33}}{3^{26}.2^7.13}=\frac{3^7}{2^7.13}\)
5 tháng 7 2017
\(\left(\dfrac{2}{3}\right)^3.\left(\dfrac{-3}{4}\right)^2.\left(-1\right)^{2013}=\dfrac{8}{27}.\dfrac{9}{16}.\left(-1\right)=-\dfrac{1}{6}\)
\(\left(\dfrac{1}{5}\right)^{15}.\left(\dfrac{1}{4}\right)^{20}=\dfrac{1}{5^{12}}.\dfrac{1}{4^{20}}=5^{-12}.4^{-20}=125^{-4}.1024^{-4}=\left(125.1024\right)^{-4}=128000^{-4}\)
5 tháng 7 2017
\(\dfrac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}=\dfrac{2^{12}.3^{10}+2^3.3.5.2^9.3^9}{2^{12}.3^{12}+2^{11}.3^{11}}=\dfrac{2^{12}.3^{10}+2^{12}.2^{10}.5}{2^{12}.3^{12}+2^{11}.3^{11}}=\dfrac{2^{12}.3^{10}\left(1+5\right)}{2^{11}.3^{11}\left(2.3+1\right)}=\dfrac{2.6}{3.7}=\dfrac{4}{7}\)
XO
10 tháng 7 2023
ĐKXĐ : \(x^4+\left(\sqrt{3}-\sqrt{2}\right).x^2-\sqrt{6}\ne0\)
\(\Leftrightarrow x\ne\sqrt[4]{2}\)
\(P=\dfrac{x^2-\sqrt{2}}{x^4+\left(\sqrt{3}-\sqrt{2}\right).x^2-\sqrt{6}}\)
\(=\dfrac{x^2-\sqrt{2}}{\left(x^4-\sqrt{2}x^2\right)+\sqrt{3}\left(x^2-\sqrt{2}\right)}\)
\(=\dfrac{x^2-\sqrt{2}}{\left(x^2+\sqrt{3}\right)\left(x^2-\sqrt{2}\right)}=\dfrac{1}{x^2+\sqrt{3}}\)
11 tháng 4 2023
1: =>4a^3+4b^3-a^3-3a^2b-3ab^2-b^3>=0
=>a^3-a^2b-ab^2+b^3>=0
=>(a+b)(a^2-ab+b^2)-ab(a+b)>=0
=>(a+b)(a-b)^2>=0(luôn đúng)
2: \(a^4+b^4=\dfrac{a^4}{1}+\dfrac{b^4}{1}>=\dfrac{\left(a^2+b^2\right)^2}{1}=\dfrac{1}{2}\left(\dfrac{a^2}{1}+\dfrac{b^2}{1}\right)^2\)
=>\(a^4+b^4>=\dfrac{1}{2}\left(\dfrac{\left(a+b\right)^2}{2}\right)^2=\dfrac{\left(a+b\right)^4}{8}\)
\(=\dfrac{\left(2^3\right)^3.\left(3^2\right)^4-2^8.\left(3^4\right)^2}{\left(2^4\right)^2.\left(3^4\right)^2+\left(2^2\right)^4.\left(3^3\right)^3}=\dfrac{2^9.3^8-2^8.3^8}{2^8.3^8+2^8.3^9}=\)
\(=\dfrac{2^8.3^8.\left(2-1\right)}{2^8.3^8.\left(1+3\right)}=\dfrac{1}{4}\)